Let $d_1$ and $d_2$ be two metrics on a non empty set $X$. Then prove that $d(x,y) = d_1 + k d_2$ is a metric.
Where $k \in \Bbb R$
My Attempt:
We know that
$d_1 + k d_2$ is a metric if $k \geq 0$ . If I take $k = 1$ then $d(x,y) = d_1(x,y) + d_2(x,y)$
So (i) $d(x,y) = d_1(x,y) + d_2(x,y) \geq 0$ $\forall x,y \in X$ (As $d_1(x,y) \geq 0$ and $d_2(x,y) \geq 0$ $\forall x,y \in X$)
(ii) $d(x,y) = 0$ iff $d_1(x,y) + d_2(x,y) = 0$ iff $d_1(x,y) = -d_2(x,y) \leq 0$ which is not possible as $d_1(x,y) \geq 0$ Therefore $d(x,y) = 0$ iff $d_1$ and $d_2$ separately equal to zero. So $d(x,y) = 0$ iff $d_1(x,y) = 0$ and $d_2(x,y) = 0$ iff $x = y$
(iii) $d(x,y) = d_1(x,y) + d_2(x,y) = d_1(y,x) + d_2(y,x) = d(x,y)$
(iv)As $d_1(x,y) \leq d_1(x,z) + d_1(z,y)$ and $d_2(x,y) \leq d_2(x,z) + d_2(z,y)$ on adding we get, $d(x,y) = d_1(x,y) + d_2(x,y) \leq d_1(x,z) + d_1(z,y) + d_2(x,z) + d_2(z,y) = d_1(x,z) + d_2(x,z) + d_1(z,y) + d_2(z,y) = d(x,z) + d(z,y)$ So $d_1 + d_2 $ is a metric on $X$
If I take $k = -1$ then
I faced a problem in each axioms except $d(x,y) = d_1(x,y) – d_2(x,y) = d_1(y,x) – d_2(y,x) = d(y,x)$ Please help me.
Best Answer
You cannot prove it, since it is false. Suppose that $d_2=d_1$ and that $k=-2$. Then $d=-d_1$, which is not a metric if $X$ has more than one point: if $x,y\in X$ and $x\ne y$, then $d(x,y)=-d_1(x,y)<0$.
However, the statment holds if $k\geqslant0$.